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We can use the identity $\tanh(x) = \frac{\sinh(2x)}{\cosh(2x)+1}$ to write $\tanh(\ln(1+x^2))$ as $$\tanh(\ln(1+x^2)) = \frac{\sinh(2\ln(1+x^2))}{\cosh(2\ln(1+x^2))+1}.$$ Now, using the identities $\sinh(2x) = 2\sinh(x)\cosh(x)$ and $\cosh(2x) = 2\cosh^2(x)-1$, we can simplify the numerator and denominator: \begin{align*} \sinh(2\ln(1+x^2)) &= 2\sinh(\ln(1+x^2))\cosh(\ln(1+x^2))
&= 2x\frac{1+x^2}{2x^2}\cdot\frac{1}{2}\left(\frac{1+x^2}{x^2}\right)^{\frac{1}{2}}
&= \frac{x(1+x^2)^{\frac{3}{2}}}{x^2+1},
\cosh(2\ln(1+x^2)) &= 2\cosh^2(\ln(1+x^2))-1
&= \frac{(1+x^2)^2}{2x^2+2}. \end{align*} Substituting these expressions back into the original equation, we get $$\tanh(\ln(1+x^2)) = \frac{x(1+x^2)^{\frac{3}{2}}}{(x^2+1)(2x^2+2)+1} = \frac{x(1+x^2)^{\frac{3}{2}}}{2x^4+4x^2+3}.$$ Thus, we have expressed $\tanh(\ln(1+x^2))$ as a rational function of $x$ and the square root of $1+x^2$. Since the square root can be expressed in terms of the real and imaginary parts of $\tanh(\ln(1+x^2))$, we can also write $\tanh(\ln(1+x^2))$ as a sum of its real and imaginary parts. However, this expression is quite complicated and not particularly illuminating.